Mathematicians Identify Threshold at Which Shapes Give Way(quantamagazine.org)
quantamagazine.org
Mathematicians Identify Threshold at Which Shapes Give Way
https://www.quantamagazine.org/mathematicians-identify-threshold-at-which-shapes-give-way-20210603/
16 comments
Quanta Magazine nails it yet again. The connection to turbulence is also very interesting.
Only question is what the C ^ x,y notation is about, as they only explained one axis. Anyone know?
Only question is what the C ^ x,y notation is about, as they only explained one axis. Anyone know?
Also hah! I saw someone once counter, "If category theory is so universally useful, tell me how it helps the study of differential equations?" The ISA "sprint" https://www.ias.edu/math/sp/hprinciple_flex_geometry_pdes referenced by the article mentions PDEs and Fukaya categories, which I hadn't heard of, so hah!
Just don't ask me to explain it :D
Just don't ask me to explain it :D
Holder spaces. https://math.ucdenver.edu/~jmandel/classes/7760f05/spaces.pd...
"""
Let Ω be an open set in Rn, 0<α≤1, and k a nonnegative integer. The (uniform) Holder spaces Ck,α(Ω) consist of functions whose k−th order derivatives are uniformly Holder continuous with exponent α in Ω. The(local) Holder spaces Ck,α(Ω) consist of functions whose k−th order derivatives are locally Holder continuous with exponent α
"""
The Holder continuity condition looks sorta like the limit definition of a derivative with an exponent on the denominator.
Also: https://en.wikipedia.org/wiki/H%C3%B6lder_condition
"""
Let Ω be an open set in Rn, 0<α≤1, and k a nonnegative integer. The (uniform) Holder spaces Ck,α(Ω) consist of functions whose k−th order derivatives are uniformly Holder continuous with exponent α in Ω. The(local) Holder spaces Ck,α(Ω) consist of functions whose k−th order derivatives are locally Holder continuous with exponent α
"""
The Holder continuity condition looks sorta like the limit definition of a derivative with an exponent on the denominator.
Also: https://en.wikipedia.org/wiki/H%C3%B6lder_condition
Thank you!! I guess in hindsight they did talk about the exponents and the iterated differentiabilty, but I didn't make the connection.
I think this was at the end of the analysis textbook too, in one of those "looking ahead future topics" sections.
I think this was at the end of the analysis textbook too, in one of those "looking ahead future topics" sections.
> The Holder continuity condition looks sorta like the limit definition of a derivative with an exponent on the denominator.
This is actually stronger than the epsilon-delta definition of continuity, as noted in the wiki article -- "For any α > 0, the condition implies the function is uniformly continuous."
Normally, you'd have something like
For all ϵ > 0, there exists δ(x, ϵ) > 0 such that
if |x - x₀| < δ, then |f(x) - f(x₀)| < ϵ.
(namely, at any given point, there's a ball of nearby points such that all their corresponding outputs are close together.)
But, with absolute continuity, δ is purely a function of ϵ, namely, that ball doesn't change size even as we move further away.
For instance, f(x)=x is absolutely continuous, while f(x)=x^2 is not. (for the former case, we can use δ = ϵ/2; for the latter case, we'd need a δ that looks something like 2ϵ·x)
In this case, since the Holder condition depends only on the distance between x and y, it automatically implies absolute continuity.
This is actually stronger than the epsilon-delta definition of continuity, as noted in the wiki article -- "For any α > 0, the condition implies the function is uniformly continuous."
Normally, you'd have something like
For all ϵ > 0, there exists δ(x, ϵ) > 0 such that
if |x - x₀| < δ, then |f(x) - f(x₀)| < ϵ.
(namely, at any given point, there's a ball of nearby points such that all their corresponding outputs are close together.)
But, with absolute continuity, δ is purely a function of ϵ, namely, that ball doesn't change size even as we move further away.
For instance, f(x)=x is absolutely continuous, while f(x)=x^2 is not. (for the former case, we can use δ = ϵ/2; for the latter case, we'd need a δ that looks something like 2ϵ·x)
In this case, since the Holder condition depends only on the distance between x and y, it automatically implies absolute continuity.
Small correction: the definition you shared is that of “uniform continuity” [1] not “absolute continuity” [2]. Absolute continuity is a slightly stronger condition where instead of providing a single (x1, x2) pair with |x1 - x2| < delta, you can now provide a finite list of pairs.
[1] https://en.m.wikipedia.org/wiki/Uniform_continuity [2] https://en.m.wikipedia.org/wiki/Absolute_continuity
[1] https://en.m.wikipedia.org/wiki/Uniform_continuity [2] https://en.m.wikipedia.org/wiki/Absolute_continuity
Ah, good catch! The wiki quote that I cited does reference uniform continuity, so the rest still holds.
(left for posterity so your comment makes sense)
(left for posterity so your comment makes sense)
Embeddings were not invented by Nash, were they?
No, but Nash was the first (AFAIK) to prove that every Riemannian manifold can be isometrically embedded in an Euclidean space of sufficiently high dimension. See, e.g., [1,2]. Previous embedding theorems (see Whitney embedding [3]) preserved topology but not necessarily geometry.
[1] https://en.wikipedia.org/wiki/Embedding#Riemannian_and_pseud...
[2] https://en.wikipedia.org/wiki/Nash_embedding_theorem
[3] https://en.wikipedia.org/wiki/Whitney_embedding_theorem
[1] https://en.wikipedia.org/wiki/Embedding#Riemannian_and_pseud...
[2] https://en.wikipedia.org/wiki/Nash_embedding_theorem
[3] https://en.wikipedia.org/wiki/Whitney_embedding_theorem
> Previous embedding theorems (see Whitney embedding [3]) preserved topology but not necessarily geometry.
The definition of Riemmannian manifold fairly straightforwardly means that every compact smooth (I think you only need C^2) manifold can be isometrically embedded in a high enough dimensional euclidean space (with the euclidean metric) just by using the fact that every riemannian manifold is locally euclidean, which means we can cover it with balls on which there are local isomorphisms, and from these we can pick a finite subcover and partition of unity that glues together for a global isometry. Basically you are gluing together local solutions to a PDE. This is something often covered in a first year course on geometry. For the details of this approach worked out, see this expository paper: https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf
To the best of my knowledge, this result doesn't even have a name attached to it (I could be wrong, but it was always presented as just a series of unnamed propositions whenever I saw this). That's not the hard part, the hard part is putting an explicit bound on the size of the target space other than "finite". So the genius of Nash was to put a strong bound on the dimension of the target space as well as the size of the target space. Rather than using general compactness and existence arguments, this requires careful arguments and innovative geometric ideas and is really a suprising result.
The definition of Riemmannian manifold fairly straightforwardly means that every compact smooth (I think you only need C^2) manifold can be isometrically embedded in a high enough dimensional euclidean space (with the euclidean metric) just by using the fact that every riemannian manifold is locally euclidean, which means we can cover it with balls on which there are local isomorphisms, and from these we can pick a finite subcover and partition of unity that glues together for a global isometry. Basically you are gluing together local solutions to a PDE. This is something often covered in a first year course on geometry. For the details of this approach worked out, see this expository paper: https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf
To the best of my knowledge, this result doesn't even have a name attached to it (I could be wrong, but it was always presented as just a series of unnamed propositions whenever I saw this). That's not the hard part, the hard part is putting an explicit bound on the size of the target space other than "finite". So the genius of Nash was to put a strong bound on the dimension of the target space as well as the size of the target space. Rather than using general compactness and existence arguments, this requires careful arguments and innovative geometric ideas and is really a suprising result.
I have to admit my memory about these things is quite hazy. About the mathematics, I do not have anything useful to add. About the history, you may be right, but my recollection seems mostly compatible with the brief history recounted in [0], which is also cited by the note you linked. The author of the note also repeats the assertion about Nash's role in the subject (of global isometric embeddings), and explains that the simpler proof presented is due to Gunther (which came later I think).
I tried to find a more detailed history, but in a few minutes' searching on the web (and flipping through the textbooks I can access from home) did not find one. Somewhat surprising given interest in the subject! Possibly this is covered in, e.g., Spivak.
[0] https://bookstore.ams.org/surv-130/12
I tried to find a more detailed history, but in a few minutes' searching on the web (and flipping through the textbooks I can access from home) did not find one. Somewhat surprising given interest in the subject! Possibly this is covered in, e.g., Spivak.
[0] https://bookstore.ams.org/surv-130/12
Perhaps I wasn't clear in my answer, but I was pointing out that just for the existence of embeddings, this follows simply from compactness arguments -- it is Proposition 3.4 in the paper I cited (not even a theorem). However this dimension depends on the finite subcover which in turn depends on (M, g). That is a completely different question that trying to find a dimension that is independent of (M, g) and only depends on the dimension of M rather that it's geometry. The AMS article you cited starts off only discussing this second question. So again, if you just want to know "can I embed a manifold", the answer is "of course you can". But if you ask, "Can I embed every manifold of dimension n into a space of dimension f(n)", then that is a much harder question, deserving of a person's name and the status of "theorem". The first example of the paper is a conjecture that f(n) = n(n+1)/2. But if you are just asking, "who showed you can embed a given manifold into a given dimension isometrically", then I maintain this follows rather simply from partitions of unity and finite subcovers, and was known pretty much at the same time as the definition of Riemannian manifold was formalized.
Correct
0.0?
There are some misleading sentences in this article.
> The C stands for continuity and the superscript zero means the curves on the embedded surface have no derivatives, not even a first.
This is not exactly true. C maps maybe have higher derivatives and maybe they don’t. Certainly, C isn’t a class of only continuous mappings that don’t have the first derivative.
Also, the example with the letter U is bad (it tries to make a point, and it fails in doing so). A U-shaped curve can be perfectly smooth at all points. It would be “problematic” if we, for example, made the letter U by “glueing” a part of the circle and two segments.
> The C stands for continuity and the superscript zero means the curves on the embedded surface have no derivatives, not even a first.
This is not exactly true. C maps maybe have higher derivatives and maybe they don’t. Certainly, C isn’t a class of only continuous mappings that don’t have the first derivative.
Also, the example with the letter U is bad (it tries to make a point, and it fails in doing so). A U-shaped curve can be perfectly smooth at all points. It would be “problematic” if we, for example, made the letter U by “glueing” a part of the circle and two segments.
Mathematicians Identify Threshold at Which Shapes Give Way [In 8 Dimensions!]